Area and Perimeter of Triangle

Perimeter of triangle

Perimeter is the length of the boundary around any closed figure.
The boundary of a triangle consists of its three sides. Therefore, sum of length of three sides of triangle is called its perimeter.

What are the units of perimeter?

Units of perimeter measurement of a triangle are taken the same as units of length of sides of the triangle.
The few examples of units of length on the side of a triangle are meter (m), centimeter (cm) etc.. So, the units of perimeter are also taken as meter (m), centimeter (cm) etc..

How to calculate perimeter of a triangle?

Perimeter of triangle ΔABC

In this ΔABC, we have
length of AB = a
length of BC = b
and length of CA = c
So, perimeter of ΔABC is sum of length of sides AB, BC and CA
i.e. perimeter of ΔABC = a + b + c

Perimeter of an equilateral triangle

Equilateral triangle ΔABC with sides length a

In an equilateral triangle, where all sides are of equal length.
i.e. AB = a, BC = a and CA = a
So, perimeter of equilateral ΔABC = a + a + a = 3a

Perimeter of isosceles triangle

Isosceles triangle ΔABC with sides length a, b and a

In isosceles triangle, where any two sides are of equal length.
i.e. AB = a, BC = a and CA = b
So, perimeter of isosceles ΔABC = a + a + b = 2a + b

Area of triangle

Area is the total amount of space occupied by any closed figure.

What are the units of area?

Area is measured in square units.
The few examples of units of area of triangle are meter² or m² read as meter square, centimeter² or cm² read as centimeter square etc.

How to calculate area of a triangle?

Area of any triangle = $\frac{1}{2}$ X base X height
Let’s understand it from following ΔABC

Area of triangle ΔABC

where, BC is base, length of BC = b
and AO is height, length of AO = h
Therefore, area of ΔABC = $\frac{1}{2}$ X b X h

Area of equilateral triangle

Area of equilateral triangle ΔABC

In above equilateral ΔABC, where all sides of triangle are equal in length, its area is calculated as:
area of ΔABC = a²
Let’s see how it is calculated?
Here, in right angle ΔAOC
a² = h² + ($\frac{\mathrm{a}}{2}$)²       by Pythagoras theorem
a² = h² +
a² – = h²
= h²
= h²
a = h
Area of Δ = $\frac{1}{2}$ X a X a
Area of Δ = a²

Area of isosceles triangle

Area of isosceles triangle ΔABC

area of isosceles ΔABC = $\frac{\text{b}\sqrt{{\mathrm{4a}}^2–b^2}}{4}$
Let’s see how it is calculated?
Here, in right angle ΔAOC
(a)² = (h)² + ($\frac{\mathrm{b}}{2}$)²           by Pythagoras theorem
a² = h² +
a² – = h²
= h²
$\sqrt{\frac{{\mathrm{4a}}^2–b^2}{4}}$ = h
$\frac{\sqrt{{\mathrm{4a}}^2–b^2}}{2}$ = h
we know, area of Δ = $\frac{1}{2}$ X b X h
∴ area of ΔABC = $\frac{1}{2}$ X b X $\frac{\sqrt{{\mathrm{4a}}^2–b^2}}{2}$
or area of ΔABC = $\frac{\text{b}\sqrt{{\mathrm{4a}}^2–b^2}}{4}$